# How do you find all the zeros of f(x) = x^3+ 8x^2- x - 8?

Mar 22, 2016

$x = - 8 , - 1 , 1$

#### Explanation:

Notice that the coefficients of each term, when added, equal $0$:

$1 + 8 - 1 - 8 = 0$

This means that $f \left(1\right) = 0$, so $1$ is a zero of the function and $x - 1$ is a factor.

Use polynomial long division or synthetic division to determine that the other two zeros can be found from:

$\frac{{x}^{3} + 8 {x}^{2} - x - 8}{x - 1} = {x}^{2} + 9 x + 8$

To find the remaining zeros of

${x}^{2} + 9 x + 8 = 0$

$\left(x + 1\right) \left(x + 8\right) = 0$

Thus, if either of these terms equals $0$,

$x + 1 = 0 \text{ "=>" } x = - 1$

$x + 8 = 0 \text{ "=>" } x = - 8$

So, the function's three zeros are located at $x = - 8 , - 1 , 1$.

We can check a graph of the function:

graph{x^3+8x^2-x-8 [-12, 4, -33.4, 85.3]}